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Given the vector v has an initial point at (0,6) and a terminal point at (2,0), find the exact value of ||v||. Answer:

Given the vector v \mathbf{v} has an initial point at (0,6) (0,6) and a terminal point at (2,0) (2,0) , find the exact value of ∥v∥ \|\mathbf{v}\| .\newlineAnswer:

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Transformations of absolute value functions: translations and reflectionsright chevron icon

Full solution

Q. Given the vector v \mathbf{v} has an initial point at (0,6) (0,6) and a terminal point at (2,0) (2,0) , find the exact value of ∥v∥ \|\mathbf{v}\| .\newlineAnswer:
  1. Distance Formula: To find the magnitude of vector vv, we need to use the distance formula, which is derived from the Pythagorean theorem. The distance formula for a vector with initial point (x1,y1)(x_1, y_1) and terminal point (x2,y2)(x_2, y_2) is:\newline∣∣v∣∣=((x2−x1)2+(y2−y1)2)||v|| = \sqrt{((x_2 - x_1)^2 + (y_2 - y_1)^2)}\newlineLet's calculate the magnitude of vector vv with the given points.
  2. Calculate Differences: First, we find the differences in the x and y coordinates:\newlineΔx=x2−x1=2−0=2\Delta x = x_2 - x_1 = 2 - 0 = 2\newlineΔy=y2−y1=0−6=−6\Delta y = y_2 - y_1 = 0 - 6 = -6
  3. Square Differences: Now, we square the differences:\newline(Δx)2=(2)2=4(\Delta x)^2 = (2)^2 = 4\newline(Δy)2=(−6)2=36(\Delta y)^2 = (-6)^2 = 36
  4. Add Squares: Next, we add the squares of the differences:\newline(Δx)2+(Δy)2=4+36=40(\Delta x)^2 + (\Delta y)^2 = 4 + 36 = 40
  5. Find Magnitude: Finally, we take the square root of the sum to find the magnitude of vector vv:∥v∥=40\|v\| = \sqrt{40}Since 4040 is not a perfect square, we can simplify the square root by factoring out perfect squares:40=4×10=4×10=210\sqrt{40} = \sqrt{4 \times 10} = \sqrt{4} \times \sqrt{10} = 2\sqrt{10}

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